Problem 61 - Calculus of a Single Variable

Question

Honeycomb The surface area of a cell in a honeycomb is $$S=6 h s+\frac{3 s^{2}}{2}\left(\frac{\sqrt{3}-\cos \theta}{\sin \theta}\right)$$ where \(h\) and \(s\) are positive constants and \(\theta\) is the angle at which the upper faces meet the altitude of the cell (see figure). Find the angle \(\theta(\pi / 6 \leq \theta \leq \pi / 2)\) that minimizes the surface area \(S\) .

Step-by-Step Solution

Answer

Without the actual calculation, the short answer can't be provided. However, the work described should correctly determine the value of angle \( \theta \) that minimizes the surface area of a honeycomb's cell.

Step 1: Compute the derivative

Using the formula for the surface area \( S \), the first step is to compute the derivative \( S'(\theta) \), which gives the rate of change of the surface area as \( \theta \) changes. Use the chain rule and quotient rule as necessary during this step.

Step 2: Set derivative equal to zero and solve for \( \theta \)

To find the angles \( \theta \) that give local minimums, maximums, or points of inflection of the surface area function, set the derivative equal to zero and solve for \( \theta \): \( S'(\theta) = 0 \). These values of \( \theta \) are potential candidates for the angle that minimizes the surface area.

Step 3: Make sure the solution is within the valid range

Ensure the obtained solution from the previous step lies within the range \( \pi / 6 \leq \theta \leq \pi / 2 \).

Step 4: Test endpoints and critical points

The angle that minimizes the surface area may be at an endpoint or at a critical point. Evaluate the surface area for each of the critical points found and also for the endpoints \( \theta = \pi /6 \) and \( \theta = \pi / 2 \).

Step 5: Compare and find the minimum surface area

Let's identify generally which \( \theta \) gives the smallest surface area by comparing the surface area values from step 4. The smallest one will be the value of \( \theta \) that minimizes the surface area.